POJ 2406 power strings KMP中next函数的应用

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23370		Accepted: 9811

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

/* 功能Function Description:     POJ-2406
   开发环境Environment:          DEV C++ 4.9.9.1
   技术特点Technique:
   版本Version:
   作者Author:                   可笑痴狂
   日期Date:                      20120813
   备注Notes:
   题意：
        找出字符串的最小循环周期从而算出最大循环次数
*/
/*
//代码一：-----简单枚举法
#include<stdio.h>
#include<string.h>
char s[1000005];

int main()
{
    int i,len,k,j,flag;
    while(scanf("%s",s+1)&&s[1]!='.')
    {
        len=strlen(s+1);
    //    printf("%d\n%c",len,s[1]);
        for(i=1;i<=len;++i)
        {
            if(len%i)
                continue;
            flag=1;
            for(k=1;flag&&k<=i;++k)
            {
                for(j=2;j<=len/i;++j)
                {
                    if(s[k]!=s[i*(j-1)+k])
                    {
                        flag=0;
                        break;
                    }
                }
            }
            if(flag)
            {
                printf("%d\n",len/i);
                break;
            }
        }
    }
    return 0;
}
*/

//代码二：-------KMP算法中next函数的性质
/*
    设Len是s的长度
    给出结论：
    如果len%(len-next[len])==0,则字符串中必存在最小循环节，且循环次数即为 len/(len-next[len])
    
    题目大意：给出一个字符串，求出是某个字符串的n次方
    解决：KMP找出字符串的循环节
    若有字符串ai...a(j+m)，若next[len+1]=m（字符串从1开始）,意味着从ai ...am 和 aj...a(j+m)相等，
    假设中间有相交部分（或者相交部分为0）aj....am，若总长度减去相交部分剩余的部分能被总长度整除，
    又因为  从ai ...am 和 aj...aj+m相等，所以除去相交部分剩余的的两段字符串相等，而且必定满足中间相交部分也能被起始字符串整除，依照这样，
    即可得出最小的循环节 

*/

#include<stdio.h>
#include<string.h>
char s[1000005];
int next[1000005];

void get_next(int len)
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else
            j=next[j];
    }

    /*
    while(i<len)              //改进算法---改进在某些情况下的缺陷  详见数据结构课本P84
    {
        if(j==-1||s[i]==s[j])
        {
            ++i;
            ++j;
            if(s[i]!=s[j])
                next[i]=j;
            else
                next[i]=next[j];
        }
        else
            j=next[j];
    }
    */
}

int main()
{
    int i,len,k,j,flag;
    while(scanf("%s",s)&&s[0]!='.')
    {
        len=strlen(s);
        get_next(len);
        if(len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
        else
            printf("1\n");
    }
    return 0;
}